Answer:
The time taken is [tex]t = 40007 sec[/tex]
Explanation:
From the question we are told that
The diameter of the egg is [tex]d_e = 5.5cm = \frac{5.5}{100} = 5.5*10^{-2}m[/tex]
The initial temperature of egg the [tex]T_e = 4.3^{o}C[/tex]
The temperature of the boiling water [tex]T_b = 100^oC[/tex]
The heat transfer coefficient is [tex]H = 800 W/m^2 \cdot K[/tex]
The final temperature is [tex]T_e_f = 74^oC[/tex]
The thermal conductivity of water is [tex]k = 0.607 W/m^oC[/tex]
The diffusivity of the egg [tex]\alpha = 0.146 * 10^{-6} m^2 /s[/tex]
Using one term approximation
We have the
[tex]\frac{T_e_f - T_b}{T_e - T_b} = Ae^{-\lambda ^2 \tau}[/tex]
The radius is [tex]r = \frac{5.5*10^ {-2}}{2} =2.75*10^{-2}m[/tex] Note that this radius is approximation to that of a real egg
Now we need to obtain the Biot number which help indicate the value of [tex]A \ and \ \lambda[/tex] to use in the above equation
The Biot number is mathematically represented as
[tex]Bi = \frac{H r}{k}[/tex]
Substituting values
[tex]Bi = \frac{800 * 2.75 *10^{-2}}{0.607}[/tex]
[tex]= 36.24[/tex]
So for this value which greater than 0.1 the coefficient [tex]\lambda_1 \ and \ A_1[/tex] is
[tex]\lambda = 3.06632[/tex]
[tex]A = 1.9942[/tex]
Substituting this into equation 1 we have
[tex]\frac{74- 100}{4.3 - 100} = 1.9942 e^{-(3.0632^2) \tau}[/tex]
[tex]0.2717= 1.9942 e^{-(3.0632^2) \tau}[/tex]
[tex]0.2717= 1.9942 e^{-9.383 \tau}[/tex]
[tex]0.13624 = e^{-9.383 \tau}[/tex]
Taking natural log of both sides
[tex]-1.993 = -9.383\ \tau[/tex]
[tex]\tau = 0.2124[/tex]
The time required for the egg to be cooked is mathematically represented as
[tex]t = \frac{\tau r^2}{\alpha }[/tex]
substituting value is
[tex]= \frac{0.2124 * 2.75 *10^{-2}}{0.146 *10^{-6}}[/tex]
[tex]t = 40007 sec[/tex]
